The magnetic field of a plane electromagnetic wave is given by $\vec B = B_0\hat i[\cos(kz - \omega t)] + B_1\hat j\cos(kz + \omega t)$, where $B_0 = 3\times10^{-5}\ \text{T}$ and $B_1 = 2\times10^{-6}\ \text{T}$. The rms value of the force experienced by a stationary charge $Q = 10^{-4}\ \text{C}$ at $z = 0$ is closest to
Answer: (D) $0.6\ \text{N}$
A stationary charge feels only the electric field. The field is the sum of two waves:
- along $+z$ with $\vec B$ along $\hat i$: $\vec E$ along $-\hat j$, amplitude $cB_0 = 9\times10^3\ \text{V/m}$
- along $-z$ with $\vec B$ along $\hat j$: $\vec E$ along $-\hat i$, amplitude $cB_1 = 600\ \text{V/m}$
At $z = 0$ both vary as $\cos\omega t$, and they are perpendicular, so the amplitude of the total field is
$$E_0 = \sqrt{(9000)^2 + (600)^2} \approx 9.02\times10^3\ \text{V/m}$$
$$F_{rms} = \frac{QE_0}{\sqrt2} = \frac{10^{-4}\times9.02\times10^3}{1.414} \approx 0.64\ \text{N} \approx 0.6\ \text{N}$$
Solution by Sreeraj P, M.Sc Physics